Question

Assume the room the tent is in will be kept about 72F. The tent is 27 x 19 x 63. Assuming the required ice would fit in the tent, I'd like to keep a few containers of that size full of ice in a freezer and swap them out when necessary, preferably once or twice a day. The ice would be on the top rack (I imagine that would be the best spot as the cold air would fall) and probably need to sit on a tray to collect condensation.

What mass of ice would I need, inside a (presumably plastic) container on a tray, to keep this tent 12F cooler? If I'm missing any necessary information, I'll update my question, or you can make assumptions based on common household conditions
Options:
A) 3.7 lbs
B) 6.2 lbs
C) 8.5 lbs
D) 10.1 lbs

Answer 1

To determine the mass of ice needed to keep the tent 12°F cooler, we can use the concept of specific heat and the formula Q = mcΔT. By calculating the heat gained by the tent and rearranging the formula, we find that the mass of ice needed is approximately 0.3267 kg (or 0.72 lbs).

Step-by-step explanation:

To determine the mass of ice needed to keep the tent 12°F cooler, we can use the concept of specific heat. The formula for specific heat is Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we know the change in temperature (12°F) and we want to find the mass of ice. We can assume that the specific heat capacity of ice is 2.09 J/g°C.

First, we convert the change in temperature from Fahrenheit to Celsius. ΔT = (12°F - 0°F) × (5/9) = 6.67°C.

Next, we can rearrange the formula to solve for mass: m = Q / (c × ΔT).

Since we want to keep the tent 12°F cooler, the amount of heat transferred is equal to the heat gained by the tent, which can be calculated using the formula Q = mcΔT. Assuming the room temperature is 72°F, the change in temperature is 12°F, and the specific heat capacity of air is approximately 1.01 J/g°C.

Plugging in the values, we get Q = mcΔT = (m × 1.01 J/g°C × 12°F) = 12.12 m.

Simplifying the equation, we now have m = Q / (c × ΔT) = 12.12 m / (2.09 J/g°C × 6.67°C) ≈ 0.3271 m.

To convert from mass to pounds, we can use the conversion factor of 1 pound = 453.592 grams. Therefore, the mass of ice needed is 0.3271 m × 453.592 g = 148.23 grams.

Answer: The mass of ice needed to keep the tent 12°F cooler is approximately 0.3267 kg (or 0.72 lbs).